12 multiple-choice questions, progressively harder.
To give f(x)=x2f(x) = x^2f(x)=x2 an inverse function, its domain can be restricted to which set?
Solution
Correct answer: D
Squaring must be made one-to-one, which means no two inputs may share an output. Keeping only x≥0x \ge 0x≥0 removes the negative twin of every input.
f(x)=x2, x≥0 ⟹ f−1(x)=xf(x) = x^2, \; x \ge 0 \;\Longrightarrow\; f^{-1}(x) = \sqrt{x}f(x)=x2,x≥0⟹f−1(x)=x
The other sets still contain both a positive and a negative input, so they are not one-to-one.
On which domain is f(x)=x2−6xf(x) = x^2 - 6xf(x)=x2−6x one-to-one and therefore invertible?
Correct answer: B
Rewrite to find the turning point: x2−6x=(x−3)2−9x^2 - 6x = (x - 3)^2 - 9x2−6x=(x−3)2−9, so the vertex is at x=3x = 3x=3. Keep one side of it.
x≥3x \ge 3x≥3
The sets x≥0x \ge 0x≥0 and x≤6x \le 6x≤6 each contain 222 and 444, which share an output, so they are not one-to-one.
Two functions satisfy f(g(x))=xf(g(x)) = xf(g(x))=x but g(f(x))≠xg(f(x)) \ne xg(f(x))=x. Are they inverses?
Correct answer: C
Inverses must undo each other from both sides, so both compositions have to return xxx.
f(g(x))=x and g(f(x))=xf(g(x)) = x \;\text{ and }\; g(f(x)) = xf(g(x))=x and g(f(x))=x
One direction alone does not make them inverses.
A spring's length is L(w)=12+0.5wL(w) = 12 + 0.5wL(w)=12+0.5w centimetres for a weight of www grams. Which function recovers the weight from a length LLL?
Solve the length rule for www.
L=12+0.5w ⟹ L−12=0.5w ⟹ w=2(L−12)L = 12 + 0.5w \;\Longrightarrow\; L - 12 = 0.5w \;\Longrightarrow\; w = 2(L - 12)L=12+0.5w⟹L−12=0.5w⟹w=2(L−12)
Let f(x)=1xf(x) = \dfrac{1}{x}f(x)=x1 and g(x)=1xg(x) = \dfrac{1}{x}g(x)=x1. What is f(g(x))f(g(x))f(g(x))?
Correct answer: A
Substitute ggg into fff and simplify the complex fraction.
f(g(x))=11x=xf(g(x)) = \frac{1}{\frac{1}{x}} = xf(g(x))=x11=x
So 1x\dfrac{1}{x}x1 is its own inverse.
Using L(w)=12+0.5wL(w) = 12 + 0.5wL(w)=12+0.5w, a spring measures 202020 centimetres. What weight is on it?
Apply the inverse w=2(L−12)w = 2(L - 12)w=2(L−12) at L=20L = 20L=20.
w=2(20−12)=2⋅8=16w = 2(20 - 12) = 2 \cdot 8 = 16w=2(20−12)=2⋅8=16
Check: L(16)=12+0.5(16)=20L(16) = 12 + 0.5(16) = 20L(16)=12+0.5(16)=20 centimetres.
Let f(x)=x2f(x) = x^2f(x)=x2 on all real numbers and g(x)=xg(x) = \sqrt{x}g(x)=x. Compute g(f(−4))g(f(-4))g(f(−4)).
Work from the inside out, remembering the square root returns the nonnegative value.
g(f(−4))=(−4)2=16=4g(f(-4)) = \sqrt{(-4)^2} = \sqrt{16} = 4g(f(−4))=(−4)2=16=4
The result is 444, not −4-4−4, because the square root cannot restore the negative sign.
For f(x)=(x−1)2f(x) = (x - 1)^2f(x)=(x−1)2 restricted to x≥1x \ge 1x≥1, find f−1(9)f^{-1}(9)f−1(9).
The inverse is f−1(x)=1+xf^{-1}(x) = 1 + \sqrt{x}f−1(x)=1+x (taking the branch with x≥1x \ge 1x≥1), so evaluate at 999.
f−1(9)=1+9=1+3=4f^{-1}(9) = 1 + \sqrt{9} = 1 + 3 = 4f−1(9)=1+9=1+3=4
Check: f(4)=(4−1)2=9f(4) = (4 - 1)^2 = 9f(4)=(4−1)2=9.
For f(x)=x2f(x) = x^2f(x)=x2 restricted to x≥0x \ge 0x≥0, find f−1(49)f^{-1}(49)f−1(49).
The inverse is f−1(x)=xf^{-1}(x) = \sqrt{x}f−1(x)=x, which takes the nonnegative root.
f−1(49)=49=7f^{-1}(49) = \sqrt{49} = 7f−1(49)=49=7
The restriction x≥0x \ge 0x≥0 is what rules out −7-7−7.
A price is first discounted to 909090 percent of its value, then a flat 333 dollar fee is added, giving the final price P(x)=0.9x+3P(x) = 0.9x + 3P(x)=0.9x+3. Which function recovers the original price xxx?
Undo the steps in reverse: subtract the fee, then undo the discount.
P=0.9x+3 ⟹ P−3=0.9x ⟹ x=P−30.9P = 0.9x + 3 \;\Longrightarrow\; P - 3 = 0.9x \;\Longrightarrow\; x = \frac{P - 3}{0.9}P=0.9x+3⟹P−3=0.9x⟹x=0.9P−3
For f(x)=x2+1f(x) = x^2 + 1f(x)=x2+1 restricted to x≥0x \ge 0x≥0, find f−1(x)f^{-1}(x)f−1(x).
Swap xxx and yyy, then solve for yyy, keeping the nonnegative root.
x=y2+1 ⟹ y2=x−1 ⟹ y=x−1x = y^2 + 1 \;\Longrightarrow\; y^2 = x - 1 \;\Longrightarrow\; y = \sqrt{x - 1}x=y2+1⟹y2=x−1⟹y=x−1
For f(x)=x+23f(x) = \dfrac{x + 2}{3}f(x)=3x+2, which gives the inverse rule and the value f−1(4)f^{-1}(4)f−1(4)?
Swap xxx and yyy, solve for yyy, then evaluate at 444.
x=y+23 ⟹ y=3x−2,f−1(4)=3(4)−2=10x = \frac{y + 2}{3} \;\Longrightarrow\; y = 3x - 2, \qquad f^{-1}(4) = 3(4) - 2 = 10x=3y+2⟹y=3x−2,f−1(4)=3(4)−2=10
Check: f(10)=10+23=4f(10) = \dfrac{10 + 2}{3} = 4f(10)=310+2=4.
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